2
\$\begingroup\$

I am currently reading chapter 12, "The Common-mode Feedback Circuit" in Analysis and Design of Analog Integrated Circuits by Gray and Hurst.

Page 815, gives the following statement for improvement of the resistive output loading of a common mode feedback (CMFB) circuit.

The following figure 12.17 is the CMFB circuit after improvement, the previous figure 12.16 is the CMFB circuit before improvement.

To avoid this resistive output loading, voltage buffers can be added between the op-amp outputs and the Rcs resistors. Source followers are used as buffers in Fig. 12.17. One potential problem is that each source follower introduces a DC offset of VGS between its input and output. To avoid a shift in the CM operating point caused by these offsets, voltage VCM can be buffered by an identical source follower so that the op-amp output voltages and VCM experience equal offsets. However, these offsets limit the op-amp output swing since each source-follower transistor that connects to an op-amp output must remain in the active region over the entire output voltage swing.

Fig12.16:

Fig12.16

Fig12.17:

Fig12.17

  1. To avoid this resistive output loading, voltage buffers can be added between the op-amp outputs and the Rcs resistors. Source followers are used as buffers in Fig. 12.17

Does this mean that the voltage sensing circuit in the previous Fig. 12.16 has too high of an output impedance? Therefore, we employ a source follower as a voltage buffer to achieve a lower output impedance for the sensing circuit?

  1. To avoid a shift in the CM operating point caused by these offsets, voltage VCM can be buffered by an identical source follower so that the op-amp output voltages and VCM experience equal offsets.

Why can we avoid a shift in the CM operating point by buffering VCM to let op-amp output voltages and VCM experience equal offsets?

  1. Why will the offset of a source-followe which is connected to the output of the opamp as a voltage buffer limit the op-amp output swing?
\$\endgroup\$

1 Answer 1

2
\$\begingroup\$
  1. When you add resistors at the output of your amplifier, you're forcing the amplifier to drive not only the "real" load (another amplifier, an ADC, etc), but also these resistors. But you can't avoid using them because CMFB is a must in a fully-diff opamp. This can be a problem because your amplifier might not have enough capability to provide current for that extra load, and as such, it can cause gain reduction, increased distortion, etc.

Adding a voltage buffer relieves the output from having to drive these resistors.

  1. you need to rephrase this question. However, it sounds like you don't understand why adding the third buffer would help compensate for the offsets introduced by the first 2. If I understood correctly, then simple answer is that the output buffers, being simple common-drain stages, shift down the output voltage by \$V_{GS}\$. Therefore, you have the following expression going into the error amplifier:

$$ V_{cm} = \frac{(V_{o1}-V_{GS})+(V_{o2}-V_{GS})}{2} $$

Assuming everything has settled, the output is simply \$V_{o,average}-V_{GS}\$. If the other (+) input of your error amplifier is simply \$V_{cm}\$, then, this error amplifier will do whatever it takes to equalize it's inputs. Since one input is at \$V_{o,average}-V_{GS}\$, it will try to get rid of that \$V_{GS}\$ term. The only way to do that is to raise the outputs driving the buffers by said voltage. Therefore, you have an offset.

Including that same \$V_{GS}\$ term at the (-) terminal is the only way out to get rid of this offset.

  1. The common-drain buffers need to comply with their \$V_{DS}\$ being larger than \$V_{GS}-V_{th}\$; you may remember this as the saturation condition for MOS transistor.

\$V_{GS}=V_{G}-V_{S}\$ and \$V_{G}\$ is now that output of your amplifier. If this large, you may drive your buffers out of saturation thus rendering your CMFB loop useless. That's why the headroom of your output amplifier is limited; unless you use a larger VDD for your buffers, thus increasing its linear range larger than that of the amplifier. Sometimes, this is possible within an IC, sometimes not.

\$\endgroup\$
4
  • \$\begingroup\$ Really appreciate of your answer,but I still have some confusions about the last answer 1:About your sentence"VGS=VG−VS and VG is now that output of your amplifier" How is VG the output of the amplifier, for the Common-drain amplifier, isn't the output $V_{o1}-V_{gs}$? \$\endgroup\$
    – Tong Su
    Feb 16 at 14:10
  • \$\begingroup\$ 2.The author said that “However, these offsets limit the op-amp output swing since each source-follower transistor that connects to an op-amp output must remain in the active region over the entire output voltage swing. " I understand that the headroom voltage of the active region of the transistor limits the overall voltage swing headroom, but how is this related to offset? \$\endgroup\$
    – Tong Su
    Feb 16 at 14:11
  • 1
    \$\begingroup\$ @TongSu No, the outputs that will be driving any next stage after this amplifier are BEFORE the buffer. These buffers are only there to drive the resistors. \$\endgroup\$
    – Designalog
    Feb 16 at 14:42
  • 1
    \$\begingroup\$ @TongSu they are related indirectly. The reasoning is as follows: 1) Resistors are loading the output too much. Use buffers. 2) Buffers introduce offset, need to compensate for them such that my outputs are not offset either. 3) But even if you compensate, you now have an extra constraint on your output swing, their \$V_{gs}\$ cannot be so big because the buffers might go into triode region. \$\endgroup\$
    – Designalog
    Feb 16 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.